Varied use of the AM-GM inequality This question appeared in the IMO some year. I have done it in 2 different ways that seem absolutely correct. Please tell me which one is right and why. I fell both are very interesting. The question is as follows:
If $x, y, z$ are positive reals, and $x+y+z=1$, find the least value of the expression :  $\frac4x + \frac9y + \frac{16}z$.
Answer 1: I can write the expression as:
$\left(\frac4x + \frac9y + \frac{16}z\right).(x+y+z)
=(4+9+16) + (4\frac{y}x + 9\frac xy) + (4\frac zx + 16\frac xz) + (16\frac yz + 9\frac zy)$
Let us term these expressions as:  $\left(4\frac{y}x + 9\frac xy\right)= t_1$ , $\left(4\frac zx + 16\frac xz\right)= t_2$ , $\left(16\frac yz + 9\frac zy\right)=t_3$
Applying the AM-GM inequality to $t_1, t_2, t_3$ we get:
$$\frac{t_1}2 ≥ 6  \implies  t_1 ≥ 12$$
$$\frac{t_2}2 ≥ 8   \implies  t_2 ≥ 16$$
$$\frac{t_3}2 ≥ 12  \implies  t_3 ≥ 24$$
Therefore:
$$\frac4x + \frac9y + \frac{16}z  ≥  4+9+16+12+16+24  =  81$$  (answer)
Answer 2: Applying the Am-GM inequality to both the set of terms:  $x, y, z$ and $\frac4x , \frac9y , \frac{16}z$, we get:
$$\frac{(x+y+z)}3   ≥  (xyz)^{\frac13}\implies \frac13 ≥  (xyz)^{\frac13}$$
[as $x+y+z = 1$]
On the other hand, applying the AM-GM inequality to $\frac4x + \frac9y + \frac{16}z = D$ (say) we get:
$$\frac D3  ≥  \left[\frac{4*9*16}{xyz}\right]^{\frac13} \implies D  ≥  [\frac{4*9*16*27}{xyz}]^{\frac13}$$
Now, we can substitute $xyz$ (put in bold) as $\frac13$, because inequality will still hold as we are increasing (or keeping same) the denominator of the lesser fraction in the inequality. Therefore, the expression becomes:
$$D  ≥  (4*9*16*27*27)^{\frac13}  =  74.88$$  (answer)
Please tell me which is right and why?
 A: The key issue is that while you may have an inequality such as $f(x) \ge c$ for $x \in [a, b]$, whether you can actually have $f(x) = c$ for some $x$.  If you can have equality, then $c$ is the minimum.  If not, it remains a lower bound. 
Assuming your workings are correct, both inequalities can simultaneously be true.  The first inequality indicates that while the second inequality may be true, it is not giving you the minimum as $~74.88$ can never be achieved. In the second case it is easy to see why - for both AM-GMs to be satisfied, you will need $x=y=z=\frac13$ and $\frac4x=\frac9y=\frac{16}z$ to hold true simultaneously, which is not possible.
BTW, an alternate method to the first inequality is Cauchy-Schwarz:
$$\left(\frac4x+\frac9y+\frac{16}z \right)=\left(\frac4x+\frac9y+\frac{16}z \right)(x+y+z)\ge (2+3+4)^2=81$$
with equality when $x:y:z=2:3:4 \implies x=\frac29, y=\frac13, z = \frac49$.  Checking equality case is a good habit, even when you are not looking for the extremum of a function.
